# How does the equation imply the algebraic multiplicity of $\lambda$ of $T$ is same as the algebraic multiplicity of $(z - \lambda)$ of $(zI - T)\$?

Let $$V$$ be a finite dimensional vector space and $$T \in \mathcal L (V).$$ Then for any $$\lambda, z \in \mathbb C$$ consider the following identity $$:$$ $$-(T - \lambda I) = (z I - T) - (z - \lambda) I.$$

This shows that $$\lambda$$ is an eigenvalue of $$T$$ if and only if $$(z - \lambda)$$ is an eigenvalue of $$(z I - T).$$ Raising both sides of the above equation to the $$\dim V$$ power and taking null spaces of both sides shows that the multiplicity of $$\lambda$$ as an eigenvalue of $$T$$ coincides with the multiplicity of $$(z - \lambda)$$ as an eigenvalue of $$(z I - T).$$ The bold faced argument is given in Sheldon Axler's book Linear Algebra Done Right which I am unable to follow. Could anyone please provide some insight on it?

Thanks for your time.

• You are basically just translating the spectrum of your matrix when you add a multiple of the identity Commented Sep 9, 2022 at 14:45

The desired result follows from the sentence above in boldface and the book's definition of the multiplicity of $$\lambda$$ as an eigenvalue of $$T$$ to be dim null $$(T - \lambda I)^{\dim V}$$.
Let $$n=\dim V.$$ The multiplicity of the eigenvalue $$\lambda$$ of $$T$$ is by definition equal to $$\dim\ker(\lambda I-T)^n.$$ We have $$\dim\ker (\lambda I-T)^n=\dim\ker [(z-\lambda)I-(zI-T)]^n$$ Therefore a number $$\lambda_0$$ is an eigenvalue with multiplicity $$k$$ of $$T$$ if and only if $$z-\lambda_0$$ is an eigenvalue with multiplicity $$k$$ of $$zI-T.$$
• Sheldon Axler proved as a corollary to the argument (I mentioned in bold faced letters) that the characteristic polynomial of $T$ is of the form $\det (\lambda I - T).$ So we can't use this to prove the result what we are supposed to. You didn't use the anywhere the author's line of reasoning. I just want you to clarify only the bold faced argument if you can. Commented Sep 9, 2022 at 14:27
• How do you know that the algebraic multiplicity of $\lambda$ equals to $\dim \text {ker} (\lambda I - T)^n\$? Commented Sep 9, 2022 at 17:36